Class 9 - Chapter 4: Describing Motion Around Us

Exercise Solution

1. My father went to a shop from home which is located at a distance of 250 m on a straight road.

Given:

• Distance between home and shop = 250 m

Total distance travelled:

Home → Shop = 250 m
Shop → Home = 250 m
Home → Shop = 250 m
Shop → Home = 250 m

Total Distance = 250 + 250 + 250 + 250

Total Distance = 1000 m

Displacement:

Initial position = Home
Final position = Home

Displacement = 0 m

Answer:

• Total Distance Travelled = 1000 m
• Displacement = 0 m

2. A student runs from the ground floor to the fourth floor and then comes down to the second floor.

Given:

• Height of each floor = 3 m

Ground Floor = 0 m
Fourth Floor = 4 × 3 = 12 m
Second Floor = 2 × 3 = 6 m

(i) Total vertical distance travelled

Ground Floor → Fourth Floor = 12 m

Fourth Floor → Second Floor = 12 − 6 = 6 m

Total Distance = 12 + 6 = 18 m

Total Vertical Distance = 18 m

(ii) Displacement

Initial Position = Ground Floor (0 m)

Final Position = Second Floor (6 m)

Displacement = 6 − 0 = 6 m upward

Displacement = 6 m upward

3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating?

Answer:

Yes, it is possible.

Acceleration depends on change in velocity. Velocity changes when either speed or direction changes. Even if the speedometer shows a constant speed, the scooter can still accelerate while taking a turn because its direction changes continuously.

4. A car starts from rest and its velocity reaches 24 m s^-1 in 6 s.

Given:

Initial velocity, u = 0 m s^-1
Final velocity, v = 24 m s^-1
Time, t = 6 s

Acceleration:

a = (v − u) / t

a = (24 − 0) / 6

a = 4 m s^-2

Acceleration = 4 m s^-2

Distance travelled:

s = ((u + v) / 2) × t

s = ((0 + 24) / 2) × 6

s = 12 × 6

s = 72 m

Distance Travelled = 72 m

5. A motorbike moving with initial velocity 28 m s^-1 comes to rest after travelling 98 m.

Given:

u = 28 m s^-1
v = 0 m s^-1
s = 98 m

Acceleration:

v² = u² + 2as

0 = (28)² + 2 × a × 98

0 = 784 + 196a

196a = −784

a = −4 m s^-2

Acceleration = −4 m s^-2

Time taken to stop:

v = u + at

0 = 28 − 4t

4t = 28

t = 7 s

Time Taken = 7 s

6. Fig. 4.27 shows position-time graphs of objects A and B.

Answer:

Yes, objects A and B have equal velocity at the point where the slopes of their position-time graphs become equal.

Since velocity is represented by the slope of a position-time graph, equal slopes indicate equal velocities.

7. Fig. 4.28 shows the change in position with time for two objects A and B.

Correct Option:

(i) and (ii)

Explanation:

• Both A and B start from the same position and end at the same position after 10 s.
• Average Velocity = Displacement / Time
• Since displacement and time are equal, average velocities are equal.
• The total distance covered by both objects is also equal; therefore average speeds are equal.

Hence, options (i) and (ii) are correct.

8. A truck driver slows down from 54 km h^-1 to 36 km h^-1 in 36 s.

Given:

Initial speed = 54 km h^-1

u = 54 × (5/18)

u = 15 m s^-1

Final speed = 36 km h^-1

v = 36 × (5/18)

v = 10 m s^-1

Time, t = 36 s

Distance travelled:

s = ((u + v) / 2) × t

s = ((15 + 10) / 2) × 36

s = (25 / 2) × 36

s = 450 m

Distance Travelled = 450 m

9. A car starts from rest and accelerates uniformly to 20 m s^-1 in 5 seconds. It then travels at 20 m s^-1 for 10 seconds and finally applies the brake (with uniform deceleration) to stop in 6 seconds. Find the total distance travelled.

Given:

Stage 1:
u = 0 m s^-1
v = 20 m s^-1
t = 5 s

Distance travelled:

s₁ = [(u + v) / 2] × t

s₁ = [(0 + 20) / 2] × 5

s₁ = 10 × 5 = 50 m

Stage 2:
Constant velocity = 20 m s^-1
Time = 10 s

s₂ = vt

s₂ = 20 × 10 = 200 m

Stage 3:
u = 20 m s^-1
v = 0 m s^-1
t = 6 s

s₃ = [(u + v) / 2] × t

s₃ = [(20 + 0) / 2] × 6

s₃ = 10 × 6 = 60 m

Total Distance = s₁ + s₂ + s₃

Total Distance = 50 + 200 + 60

Total Distance = 310 m

10. A bus is travelling at 36 km h^-1 when the driver sees an obstacle 30 m ahead.

Given:

Initial speed = 36 km h^-1

u = 36 × (5/18)

u = 10 m s^-1

Reaction time = 0.5 s

Deceleration = 2.5 m s^-2

Distance travelled during reaction time:

s₁ = ut

s₁ = 10 × 0.5

s₁ = 5 m

Braking distance:

v² = u² + 2as

0 = (10)² + 2(-2.5)s

0 = 100 - 5s

5s = 100

s = 20 m

Total stopping distance:

= Reaction distance + Braking distance

= 5 + 20

= 25 m

Obstacle distance = 30 m

Since 25 m < 30 m,

the bus will stop before reaching the obstacle.

11. A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.

Answer:

Rest and motion are relative concepts.

An object kept on the Earth is at rest with respect to the Earth and nearby surroundings. However, with respect to the Sun, the object is moving because the Earth itself revolves around the Sun and rotates on its axis.

Therefore, an object on Earth can be considered at rest in one frame of reference and moving in another.

12. Velocity–time graph of a cyclist (Fig. 4.30)

(i) Constant velocity region:

From 20 s to 100 s, the graph is horizontal at 3 m s^-1. This portion represents constant velocity.

(ii) Velocity decreasing region:

From 100 s to 120 s, the graph slopes downward. This portion represents decreasing velocity.

Displacement in 120 s:

Displacement = Area under velocity–time graph

Area of first triangle:

A₁ = (1/2) × 20 × 3 = 30 m

Area of rectangle:

A₂ = 80 × 3 = 240 m

Area of trapezium:

A₃ = (1/2)(3 + 2) × 20

A₃ = 50 m

Total displacement:

= 30 + 240 + 50

= 320 m

Average acceleration:

Initial velocity = 0 m s^-1

Final velocity = 2 m s^-1

Average acceleration

a = (v - u) / t

a = (2 - 0) / 120

a = 1/60 m s^-2 ≈ 0.0167 m s^-2

13. A girl is preparing for her first marathon by running on a straight road. Estimate the distance she ran based on Fig. 4.31.

Answer:

Distance travelled = Area under the velocity–time graph.

Using the graph:

• 0–2 h: Average velocity = (7 + 7.5)/2 = 7.25 km h^-1
• Distance = 7.25 × 2 = 14.5 km

• 2–6 h: Average velocity = (7.5 + 6.5)/2 = 7 km h^-1
• Distance = 7 × 4 = 28 km

Total distance ≈ 14.5 + 28

Total distance ≈ 42.5 km

14. A car continues to move with a constant velocity of 6 m s^-1 for 2 minutes and then accelerates at 1 m s^-2 for 6 seconds.

First stage:

v = 6 m s^-1

t = 2 min = 120 s

s₁ = vt

s₁ = 6 × 120

s₁ = 720 m

Second stage:

u = 6 m s^-1

a = 1 m s^-2

t = 6 s

s₂ = ut + (1/2)at²

s₂ = (6 × 6) + (1/2 × 1 × 36)

s₂ = 36 + 18

s₂ = 54 m

Total displacement:

= 720 + 54

= 774 m

15. Two cars A and B start from rest.

Car A:

u = 0

v = 5 m s^-1

t = 5 s

Acceleration:

a = (5 - 0)/5 = 1 m s^-2

Displacement:

s = (1/2)at²

s = (1/2)(1)(5)²

s = 12.5 m

Car B:

u = 0

v = 3 m s^-1

t = 10 s

Acceleration:

a = (3 - 0)/10

a = 0.3 m s^-2

Displacement:

s = (1/2)(0.3)(10)²

s = 15 m

The velocity–time graphs are straight lines starting from the origin. Car A reaches (5 s, 5 m s^-1) and Car B reaches (10 s, 3 m s^-1).

16. Minute hand of a clock from 6 PM to 7:30 PM

Given:

Radius = 7 cm

Time interval = 1.5 h

The minute hand completes:

1.5 revolutions

(i) Distance travelled

Distance = 1.5 × Circumference

= 1.5 × 2πr

= 1.5 × 2 × (22/7) × 7

= 66 cm

Distance travelled = 66 cm

(ii) Displacement

After 1.5 revolutions, the tip reaches the diametrically opposite point.

Displacement = Diameter

= 2 × 7

Displacement = 14 cm

(iii) Speed

Speed = Distance / Time

= 66 / 90

= 0.733 cm min^-1

(iv) Velocity

Velocity = Displacement / Time

= 14 / 90

= 0.156 cm min^-1

Direction is along the straight line joining the initial and final positions.